Transcription
Solutions Manual forThermodynamics and ChemistrySecond EditionbyHoward DeVoeAssociate Professor of Chemistry EmeritusUniversity of Maryland, College Park, Marylandhdevoe@umd.eduCopyright 2020 by Howard DeVoeThis work is licensed under a Creative Commons Attribution 4.0 International 0/
ContentsPreface31Introduction42Systems and Their Properties53The First Law84The Second Law165Thermodynamic Potentials196The Third Law and Cryogenics247Pure Substances in Single Phases268Phase Transitions and Equilibria of Pure Substances369Mixtures4110Electrolyte Solutions5511Reactions and Other Chemical Processes5812Equilibrium Conditions in Multicomponent Systems7713The Phase Rule and Phase Diagrams9414Galvanic Cells104
PrefaceThis manual contains detailed solutions to the problems appearing at the end of each chapterof the text Thermodynamics and Chemistry.Each problem printed in the text is reproduced in this manual, followed by a worked-outsolution. If a figure or table accompanies a problem in the text, it is also reproduced here.Included within a solution may be an additional figure or table that does not appear in thetext. All figures, tables, and footnotes in this manual are numbered consecutively (Figure 1,Figure 2, etc.) and so do not agree with the numbering in the text.In most cases of a numerical calculation involving physical quantities, the setup in thismanual shows the values of given individual physical quantities expressed in SI base unitsand SI derived units, without prefixes. The result of the calculation is then expressed in SIbase units and SI derived units appropriate to the physical quantity being evaluated. Sincethe factors needed to convert the units of the given quantities to the units of the calculatedquantity all have numerical values of unity when this procedure is followed, the conversionfactors are not shown.Of course, the solution given in this manual for any particular problem is probably notthe only way the problem can be solved; other solutions may be equally valid.
4Chapter 1 Introduction1.1 Consider the following equations for the pressure of a real gas. For each equation, find thedimensions of the constants a and b and express these dimensions in SI units.(a) The Dieterici equation:pDRT e .an VRT /.V n/ bSolution:Since an VRT is a power, it is dimensionless and a has the same dimensions asVRT n. These dimensions are volume energy/amount2 , expressed in m3 J mol 2 . b hasthe same dimensions as V n, which are volume/amount expressed in m3 mol 1 .(b) The Redlich–Kwong equation:pDRT.V n/ ban2T 1 2 V .V C nb/Solution:The term an2 T 1 2 V .V C nb/ has the same dimensions as p, so a has the samedimensions as T 1 2 V 2 pn 2 . The SI units are K1 2 m6 Pa mol 2 . b has the samedimensions as V n, which are volume/amount expressed in m3 mol 1 .
5Chapter 2 Systems and Their Properties2.1 Let X represent the quantity V 2 with dimensions .length/6 . Give a reason that X is or is notan extensive property. Give a reason that X is or is not an intensive property.Solution:X is not an extensive property because it is not additive: .V ’ /2 C .V “ /2 .V ’ C V “ /2 (e.g.,12 C 12 22 ).X is not an intensive property because it is dependent on volume.2.2 Calculate the relative uncertainty (the uncertainty divided by the value) for each of the measurement methods listed in Table 2.2 on page 38, using the typical values shown. For each ofthe five physical quantities listed, which measurement method has the smallest relative uncertainty?Solution:Mass:analytical balance, 0:1 10 3 g 100 g D 1 10 6micro balance, 0:1 10 6 g 20 10 3 g D 5 10 6Volume:pipet, 0:02 ml 10 mL D 2 10 3volumetric flask, 0:3 10 3 L 1 L D 3 10 4Density:pycnometer, 2 10 3 g mL 1 1 g mL 1 D 2 10 3magnetic float densimeter, 0:1 10 3 g mL 1 1 g mL 1 D 1 10 4Pressure:manometer or barometer, 0:001 Torr 760 Torr D 1 10 6diaphragm gauge, 1 Torr 100 Torr D 1 10 2Temperature:gas thermometer, 0:001 K 10 K D 1 10 4mercury thermometer, 0:01 K 300 K D 3 10 5platinum resistance thermometer, 0:0001 K 300 K D 3 10 7optical pyrometer, 0:03 K 1300 K D 2 10 5The measurement of temperature with a platinum resistance thermometer has the least relativeuncertainty, and the measurement of pressure with a diaphragm gauge has the greatest. Foreach physical quantity, the measurement method with smallest relative uncertainty isunderlined in the preceding list.2.3 Table 1 on the next page lists data obtained from a constant-volume gas thermometer containingsamples of varying amounts of helium maintained at a certain fixed temperature T2 in the gasbulb.1 The molar volume Vm of each sample was evaluated from its pressure in the bulb at areference temperature of T1 D 7:1992 K, corrected for gas nonideality with the known valueof the second virial coefficient at that temperature.Use these data and Eq. 2.2.2 on page 34 to evaluate T2 and the second virial coefficient of helium at temperature T2 . (You can assume the third and higher virial coefficients are negligible.)Solution:With the third and higher virial coefficients set equal to zero, Eq. 2.2.2 becomes BpVm D RT 1 CVm1 Ref.[13].
6Table 1 Helium at a fixed temperature.1 Vm / 102 mol m3.p2 Vm R/ 4472.62282.61622.59652.57902.55862.8.p2 Vm R/ Kbcbc2:7bcbcbcbcbcbc2:6bcbcbc2:501234.1 Vm / 102 mol m563Figure 1According to this equation, a plot of p2 Vm R versus 1 Vm should be linear with an interceptat 1 Vm D0 equal to T2 and a slope equal to BT2 . The plot is shown in Fig. 1. A least-squaresfit of the data to a first-order polynomial yields an intercept of 2:7478 K and a slope of3:659 10 4 K m3 mol 1 . The temperature and second virial coefficient therefore have thevaluesT2 D 2:7478 KBD3:659 10 4 K m3 mol2:7478 K1D1:332 104m3 mol12.4 Discuss the proposition that, to a certain degree of approximation, a living organism is a steadystate system.Solution:The organism can be treated as being in a steady state if we assume that its mass is constant
7and if we neglect internal motion. Matter enters the organism in the form of food, water, andoxygen; waste matter and heat leave the system.2.5 The value of U for the formation of one mole of crystalline potassium iodide from its elements at 25 ı C and 1 bar is 327:9 kJ. Calculate m for this process. Comment on thefeasibility of measuring this mass change.Solution: m D U c 2 D327:9 103 J .2:998 108 m s1 2/ D3:648 1012kgThis mass change is much less than the uncertainty of a microbalance (Table 2.2), which doesnot even have the capacity to weigh one mole of KI—so it is hopeless to try to measure thismass change.
8Chapter 3 The First Law3.1 Assume you have a metal spring that obeys Hooke’s law: F D c.l l0 /, where F is the forceexerted on the spring of length l, l0 is the length of the unstressed spring, and c is the springconstant. Find an expression for the work done on the spring when you reversibly compress itfrom length l0 to a shorter length l 0 .Solution:Z l0ZwDF dl D cl0l0l0.ll0 / dl D1c.l2ˇl 0l0 /2 ˇl D 12 c.l 00l0 /2airwaterFigure 23.2 The apparatus shown in Fig. 2 consists of fixed amounts of water and air and an incompressiblesolid glass sphere (a marble), all enclosed in a rigid vessel resting on a lab bench. Assume themarble has an adiabatic outer layer so that its temperature cannot change, and that the walls ofthe vessel are also adiabatic.Initially the marble is suspended above the water. When released, it falls through the air intothe water and comes to rest at the bottom of the vessel, causing the water and air (but notthe marble) to become slightly warmer. The process is complete when the system returns toan equilibrium state. The system energy change during this process depends on the frame ofreference and on how the system is defined. Esys is the energy change in a lab frame, and U is the energy change in a specified local frame.For each of the following definitions of the system, give the sign (positive, negative, or zero)of both Esys and U , and state your reasoning. Take the local frame for each system to be acenter-of-mass frame.Solution:Because q is zero in each part of this problem, Esys is equal to wlab and U is equal to w. 2set equal to zero: U Esys D wlab w D mg zcmWe can use Eq. 3.1.4 with vcmor Esys D U C mg zcm .(a) The system is the marble.Solution: U is zero, because the state of the system is unchanged. Esys is negative, because in the lab frame the marble does work on the water (page 83).This can also be deduced using Esys D U C mg zcm and the fact that U is zeroand zcm is negative.
9(b) The system is the combination of water and air.Solution: U is positive, because the system’s temperature increases at constant volume. Esys is positive, because both U and zcm are positive (the center of gravity of thewater rises when the marble enters the water). This can also be deduced by consideringthat the net force exerted by the sinking marble on the water and the displacement of theboundary at the marble are in the same direction (downward).(c) The system is the combination of water, air, and marble.Solution: Esys is zero, because wlab is zero (there is no displacement of the system boundary inthe lab frame). U is positive, because Esys is zero and zcm is negative. This can also be deducedfrom the fact that U is an extensive property, so that U for this system is equal to thesum of the internal energy change of the marble and the internal energy change of thewater and air. In parts (a) and (b) these changes were found to be zero and positive,respectively.porousplugpistonp D 3:00 barV D 0:500 m3T D 300:0 KgasText D 300:0 Kpext D 1:00 barFigure 33.3 Figure 3 shows the initial state of an apparatus consisting of an ideal gas in a bulb, a stopcock,a porous plug, and a cylinder containing a frictionless piston. The walls are diathermal, andthe surroundings are at a constant temperature of 300:0 K and a constant pressure of 1:00 bar.When the stopcock is opened, the gas diffuses slowly through the porous plug, and the pistonmoves slowly to the right. The process ends when the pressures are equalized and the pistonstops moving. The system is the gas. Assume that during the process the temperature throughout the system differs only infinitesimally from 300:0 K and the pressure on both sides of thepiston differs only infinitesimally from 1:00 bar.(a) Which of these terms correctly describes the process: isothermal, isobaric, isochoric,reversible, irreversible?Solution:The process is isothermal and irreversible, but not isobaric, isochoric, or reversible. Notethat the pressure gradient across the porous plug prevents intermediate states of theprocess from being equilibrium states, and keeps the process from being reversible; thereis no infinitesimal change that can reverse the motion of the piston.(b) Calculate q and w.Solution:Because T is constant and the gas is ideal, the relation p1 V1 D p2 V2 holds, and the finalvolume is found from
10V2 D.3:00 bar/.0:500 m3 /p 1 V1DD 1:50 m3p21:00 barThe work must be calculated from the pressure at the moving portion of the boundary(the inner surface of the piston); this is a constant pressure of 1:00 bar:Z V2wDp dV D p.V2 V1 / D .1:00 105 Pa/.1:50 0:500/m3V1D1:00 105 Jq D UwD0w D 1:00 105 J3.4 Consider a horizontal cylinder-and-piston device similar to the one shown in Fig. 3.5 onpage 72. The piston has mass m. The cylinder wall is diathermal and is in thermal contactwith a heat reservoir of temperature Text . The system is an amount n of an ideal gas confinedin the cylinder by the piston.The initial state of the system is an equilibrium state described by p1 and T D Text . Thereis a constant external pressure pext , equal to twice p1 , that supplies a constant external forceon the piston. When the piston is released, it begins to move to the left to compress the gas.Make the idealized assumptions that (1) the piston moves with negligible friction; and (2) thegas remains practically uniform (because the piston is massive and its motion is slow) and hasa practically constant temperature T D Text (because temperature equilibration is rapid).(a) Describe the resulting process.Solution:The piston will oscillate; the gas volume will change back and forth between the initialvalue V1 and a minimum value V2 .(b) Describe how you could calculate w and q during the period needed for the piston velocityto become zero again.Solution:The relation between V1 and V2 is found by equating the work done on the gas by thepiston, nRT ln.V2 V1 /, to the work done on the piston by the external pressure,pext .V2 V1 /, where pext is given by pext D 2p1 D 2nRT V1 . The result isV2 D 0:2032V1 , w D 1:5936nRT , q D w D 1:5936nRT .(c) Calculate w and q during this period for 0:500 mol gas at 300 K.Solution:w D .1:5936/.0:500 mol/.8:3145 J Kq D w D 1:99 103 J.1mol1/.300 K/ D 1:99 103 J,3.5 This problem is designed to test the assertion on page 60 that for typical thermodynamic processes in which the elevation of the center of mass changes, it is usually a good approximationto set w equal to wlab . The cylinder shown in Fig. 4 on the next page has a vertical orientation,so the elevation of the center of mass of the gas confined by the piston changes as the pistonslides up or down. The system is the gas. Assume the gas is nitrogen (M D 28:0 g mol 1 ) at300 K, and initially the vertical length l of the gas column is one meter. Treat the nitrogen asan ideal gas, use a center-of-mass local frame, and take the center of mass to be at the midpoint of the gas column. Find the difference between the values of w and wlab , expressed asa percentage of w, when the gas is expanded reversibly and isothermally to twice its initialvolume.Solution:Use Eq. 3.1.4: wwlab D1m 22vcm mg zcm .
11lgasFigure 4 2 vcmis zero, and zcm isl12l22D2l12l12D 21 l1 ; therefore wwlab D1mgl1 .2From Eq. 3.5.1, which assumes the local frame is a lab frame:Vwlab D nRT ln 2 D nRT ln 2.V1Use these relations to obtain w D wlabwwlabDwDnRT1mgl12ln 2 12 mgl1D1mgl12DnRT ln 21mgl1 .211D2nRT ln 22RT ln 2C1C1mgl1Mgl11D 7:9 10.2/.8:3145 J K mol 1 /.300 K/.ln 2/C1.28:0 10 3 kg mol 1 )(9.81 m s 2 )(1 m)51which is 0:0079%.vacuum weightidealgashFigure 53.6 Figure 5 shows an ideal gas confined by a frictionless piston in a vertical cylinder. The systemis the gas, and the boundary is adiabatic. The downward force on the piston can be varied bychanging the weight on top of it.
12(a) Show that when the system is in an equilibrium state, the gas pressure is given by p Dmgh V where m is the combined mass of the piston and weight, g is the acceleration offree fall, and h is the elevation of the piston shown in the figure.Solution:The piston must be stationary in order for the system to be in an equilibrium state.Therefore the net force on the piston is zero: pAs mg D 0 (where As is thecross-section area of the cylinder). This gives p D mg As D mg .V h/ D mgh V .(b) Initially the combined mass of the piston and weight is m1 , the piston is at height h1 , andthe system is in an equilibrium state with conditions p1 and V1 . The initial temperatureis T1 D p1 V1 nR. Suppose that an additional weight is suddenly placed on the piston,so that m increases from m1 to m2 , causing the piston to sink and the gas to be compressed adiabatically and spontaneously. Pressure gradients in the gas, a form of friction,eventually cause the piston to come to rest at a final position h2 . Find the final volume,V2 , as a function of p1 , p2 , V1 , and CV . (Assume that the heat capacity of the gas, CV ,is independent of temperature.) Hint: The potential energy of the surroundings changesby m2 g h; since the kinetic energy of the piston and weights is zero at the beginningand end of the process, and the boundary is adiabatic, the internal energy of the gas mustchange by m2 g h D m2 g V As D p2 V .Solution:There are two expressions for U : U D CV .T2T1 /and U Dp2 .V2V1 /Equate the two expressions and substitute T1 D p1 V1 nR and T2 D p2 V2 nR:.CV nR/.p2 V2p 1 V1 / DSolve for V2 : V2 Dp2 .V2V1 /CV p1 C nRp2Vp2 .CV C nR/ 1(c) It might seem that by making the weight placed on the piston sufficiently large, V2 couldbe made as close to zero as desired. Actually, however, this is not the case. Find expressions for V2 and T2 in the limit as m2 approaches infinity, and evaluate V2 V1 in thislimit if the heat capacity is CV D .3 2/nR (the value for an ideal monatomic gas at roomtemperature).Solution:Since p2 is equal to m2 g As , p2 must approach 1 as m2 approaches 1. In theexpression for V2 , the term CV p1 becomes negligible as p2 approaches 1; then p2cancels from the numerator and denominator givingV2 !nRVCV C nR 1The relation T2 D p2 V2 nR shows that with a finite limiting value of V2 , T2 mustapproach 1 as p2 does. If CV equals .3 2/nR, then V2 V1 approaches 2 5 D 0:4.3.7 The solid curve in Fig. 3.7 on page 80 shows the path of a reversible adiabatic expansion orcompression of a fixed amount of an ideal gas. Information about the gas is given in the figurecaption. For compression along this path, starting at V D 0:3000 dm3 and T D 300:0 K andending at V D 0:1000 dm3 , find the final temperature to 0:1 K and the work.Solution:CV D nCV;m D n.1:500R/ D .0:0120 mol/.1:500/.8:3145 J K1mol1/ D 0:1497 J K1
13T2 D T1 V1V2 nR CVw D CV .T2D .300:0 K/.3:000/.1 1:500/ D 624:0 KT1 / D .0:1497 J K1/.624:0 K300:0 K/ D 48:5 Jp D 3:00 barV D 0:500 m3T D 300:0 KpD0V D 1:00 m3gasvacuumText D 300:0 KFigure 63.8 Figure 6 shows the initial state of an apparatus containing an ideal gas. When the stopcock isopened, gas passes into the evacuated vessel. The system is the gas. Find q, w, and U underthe following conditions.(a) The vessels have adiabatic walls.Solution:q D 0 because the process is adiabatic; w D 0 because it is a free expansion; therefore, U D q C w D 0.(b) The vessels have diathermal walls in thermal contact with a water bath maintained at300: K, and the final temperature in both vessels is T D 300: K.Solution: U D 0 because the gas is ideal and the final and initial temperatures are the same;w D 0 because it is a free expansion; therefore q D U w D 0.3.9 Consider a reversible process in which the shaft of system A in Fig. 3.11 makes one revolutionin the direction of increasing #. Show that the gravitational work of the weight is the same asthe shaft work given by w D mgr #.Solution:The circumference of the shaft at the point where the cord is attached is 2 r. When the shaftmakes one revolution in the direction of increasing #, a length of cord equal to 2 r becomeswrapped around the shaft, and the weight rises by a distance z D 2 r. The gravitationalwork, ignoring the buoyant force of the air, is w D mg z D mgr.2 /, which is the same asthe shaft work mgr # with # D 2 .3.10 This problem guides you through a calculation of the mechanical equivalent of heat using datafrom one of James Joule’s experiments with a paddle wheel apparatus (see Sec. 3.7.2). Theexperimental data are collected in Table 2 on the next page.In each of his experiments, Joule allowed the weights of the apparatus to sink to the floortwenty times from a height of about 1:6 m, using a crank to raise the weights before eachdescent (see Fig. 3.14 on page 89). The paddle wheel was engaged to the weights through theroller and strings only while the weights descended. Each descent took about 26 seconds, andthe entire experiment lasted 35 minutes. Joule measured the water temperature with a sensitivemercury-in-glass thermometer at both the start and finish of the experiment.
14Table 2 Data for Problem 3.10. The values are from Joule’s 1850 paper a andhave been converted to SI units.Properties of the paddle wheel apparatus:combined mass of the two lead weights . . . . . . . . . . . . . . . .mass of water in vessel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .mass of water with same heat capacityas paddle wheel, vessel, and lidb . . . . . . . . . . . . . . . . . . .Measurements during the experiment:number of times weights were wound up and released . . .change of elevation of weights during each descent . . . . .final downward velocity of weights during descent . . . . . .initial temperature in vessel . . . . . . . . . . . . . . . . . . . . . . . . . . .final temperature in vessel . . . . . . . . . . . . . . . . . . . . . . . . . . . .mean air temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .a Ref.26:3182 kg6:04118 kg0:27478 kg201:5898 m0:0615 m s288:829 K289:148 K289:228 K1[91], p. 67, experiment 5.from the masses and specific heat capacities of the materials.b CalculatedFor the purposes of the calculations, define the system to be the combination of the vessel, itscontents (including the paddle wheel and water), and its lid. All energies are measured in alab frame. Ignore the small quantity of expansion work occurring in the experiment. It helpsconceptually to think of the cellar room in which Joule set up his apparatus as being effectivelyisolated from the rest of the universe; then the only surroundings you need to consider for thecalculations are the part of the room outside the system.(a) Calculate the change of the gravitational potential energy Ep of the lead weights duringeach of the descents. For the acceleration of free fall at Manchester, England (whereJoule carried out the experiment) use the value g D 9:813 m s 2 . This energy changerepresents a decrease in the energy of the surroundings, and would be equal in magnitudeand opposite in sign to the stirring work done on the system if there were no other changesin the surroundings.Solution: Ep D mg h D .26:3182 kg/.9:813 m s2/. 1:5898 m/ D410:6 J(b) Calculate the kinetic energy Ek of the descending weights just before they reached thefloor. This represents an increase in the energy of the surroundings. (This energy wasdissipated into thermal energy in the surroundings when the weights came to rest on thefloor.)Solution:Ek D .1 2/mv 2 D .1 2/.26:3182 kg/.0:0615 m sec1 2/ D 0:0498 J(c) Joule found that during each descent of the weights, friction in the strings and pulleysdecreased the quantity of work performed on the system by 2:87 J. This quantity represents an increase in the thermal energy of the surroundings. Joule also considered theslight stretching of the strings while the weights were suspended from them: when theweights came to rest on the floor, the tension was relieved and the potential energy of thestrings changed by 1:15 J. Find the total change in the energy of the surroundings duringthe entire experiment from all the effects described to this point. Keep in mind that theweights descended 20 times during the experiment.Solution: Esur D .20/ . 410:6 C 0:0498 C 2:871:15/ J D8176:6 J
15(d) Data in Table 2 show that change of the temperature of the system during the experimentwas T D .289:148 288:829/ K D C0:319 KThe paddle wheel vessel had no thermal insulation, and the air temperature was slighterwarmer, so during the experiment there was a transfer of some heat into the system. Froma correction procedure described by Joule, the temperature change that would have occurred if the vessel had been insulated is estimated to be C0:317 K.Use this information together with your results from part (c) to evaluate the work neededto increase the temperature of one gram of water by one kelvin. This is the “mechanicalequivalent of heat” at the average temperature of the system during the experiment. (Asmentioned on p. 87, Joule obtained the value 4:165 J based on all 40 of his experiments.)Solution: Esys D Esur D 8176:6 J8176:6 J.6:04118 kg C 0:27478 kg/.103 g kg1/.0:317 K/D 4:08 J g1K13.11 Refer to the apparatus depicted in Fig. 3.1 on page 61. Suppose the mass of the external weightis m D 1:50 kg, the resistance of the electrical resistor is Rel D 5:50 k , and the accelerationof free fall is g D 9:81 m s 2 . For how long a period of time will the external cell need tooperate, providing an electric potential difference j j D 1:30 V, to cause the same change inthe state of the system as the change when the weight sinks 20:0 cm without electrical work?Assume both processes occur adiabatically.Solution:The value of U is the same in both processes.Sinking of weight: U D w D mg h D .1:50 kg/.9:81 m s 2 /.20:0 10Electrical work: U D w D I 2 Rel t D IRel (Ohm’s law);therefore, U D . /2 t Rel .tDRel U.5:50 103 /.2:943 J/DD 9:58 103 s. /2.1:30 V/22m/ D 2:943 J
16Chapter 4 The Second Law4.1 Explain why an electric refrigerator, which transfers energy by means of heat from the coldfood storage compartment to the warmer air in the room, is not an impossible “Clausius device.”Solution:In addition to heat transfer, there is consumption of electrical energy from the surroundings.The refrigerator does not fit the description of the device declared by the Clausius statementof the second law to be impossible; such a device would produce no other effect but thetransfer of energy by means of heat from a cooler to a warmer body.4.2 A system consisting of a fixed amount of an ideal gas is maintained in thermal equilibriumwith a heat reservoir at temperature T . The system is subjected to the following isothermalcycle:1. The gas, initially in an equilibrium state with volume V0 , is allowed to expand into avacuum and reach a new equilibrium state of volume V 0 .2. The gas is reversibly compressed from V 0 to V0 .HHFor this cycle, find expressions or values for w, ¶q T , and dS .Solution:Step 1: w D 0 because it is a free expansion; U D 0 because it is an ideal gas andisothermal; therefore q D U w D 0.Step 2: w D nRT ln.V0 V 0 / (Eq. 3.5.1); U D 0 because it is an ideal gas and isothermal;0therefore q D U w D nRT ln.VH 0 V /.H0Overall: w D nRT ln.V V/;¶q TD Hq T D nR ln.V0 V 0 /; dS D 0 because S is aH0state function. Note that dS is greater than ¶q T because of the irreversible expansionstep, in agreement with the mathematical statement of the second law.4.3 In an irreversible isothermal process of a closed system:(a) Is it possible for S to be negative?Solution:Yes. Provided S is less negative than q T , there is no violation of the second law.(b) Is it possible for S to be less than q T ?Solution:According to the second law, no.4.4 Suppose you have two blocks of copper, each of heat capacity CV D 200:0 J K 1 . Initially oneblock has a uniform temperature of 300:00 K and the other 310:00 K. Calculate the entropychange that occurs when you place the two blocks in thermal contact with one another andsurround them with perfect thermal insulation. Is the sign of S consistent with the secondlaw? (Assume the process occurs at constant volume.)Solution:Since the blocks have equal heat capacities, a given quantity of heat transfer from the warmerto the cooler block causes temperature changes that are equal in magnitude and of oppositesigns. The final equilibrium temperature is 305:00 K, the average of the initial values.When the temperature of one of the blocks changes reversibly from T1 to T2 , the entropychange is
17 S DZ¶qDTZT2T1CV dTTD CV ln 2TT1305:00 KD 3:306 J K 1300:00 K305:00 KWarmer block: S D 200:0 J K 1 lnD 3:252 J K310:00 KCooler block: S D 200:0 J K1lnTotal entropy change: S D 3:306 J K113:252 J K1D 0:054 J K1The sign of S is positive as predicted by the second law for an irreversible process in anisolated system.4.5 Refer to the apparatus shown in Figs. 3 on page 9 and 6 on page 13 and described in Probs. 3.3and 3.8. For bothR systems, evaluate S for the process that results from opening the stopcock.Also evaluate ¶q Text for both processes (for the apparatus in Fig. 6, assume the vessels haveadiabatic walls). Are your results consistent with the mathematical statement of the secondlaw?Solution:The initial states of the ideal gas are the same in both processes, and the final states are alsothe same. Therefore the value of S is the same for both processes. Calculate S for areversible isothermal expansion of the ideal gas from the initial to the final volume: qwV2p VV2 S DDD nR lnD 1 1 lnTTV1TV1 .3:00 105 Pa/.0:500 m3 /1:50 m3Dln300: K0:500 m3D 549 J K1RUse the values of q found in Probs. 3.3 and 3.8 to evaluate ¶q Text :Forplug,R the expansion through the porous¶q Text D q T D 1:00 105 J 300: K D R333 J K 1 .For the expansion into the evacuated Rvessel, ¶q Text D q T D 0.In both processes S is greater than ¶q Text , consistent with the second law.bmwaterFigure 74.6 Figure 7 shows the walls of a rigid thermally-insulated box (cross hatching). The system is thecontents of this box. In the box is a paddle wheel immersed in a container of water, connected
18by a cord and pulley to a weight of mass m. The weight rests on a stop located a distance habove the bottom of the box. Assume the heat capacity of the system, CV , is independent oftemperature. Initially the system is in an equilibrium state at temperature T1 . When the stopis removed, the weight irreversibly sinks to the bottom of the box, causing the paddle wheelto rotate in the water. Eventually the system reaches a final equilibrium state with thermalequilibrium. Describe a reversible process with the same entropy change as this irreversibleprocess, and derive a formula for S in terms of m, h, CV , and T1 .Solution:When the stop is removed, the system is an isolated system of constant internal energy. Toreversibly change the system between the same initial and final states as the irreversibleprocess, reversibly lower the weight with gravitational work w D mgh, then let reversibleheat q D CV .T2 T1 / enter the system. Since U is zero, the sum of q and w must be zero:CV .T2T1 /mgh D 0T2 D T1 CmghCVThe entropy change of the reversible process is ZZ T2¶qdTT2mgh S DD CVD CV lnD CV ln 1 CTbTT1CV T1T1This is also the entropy change
Chapter 1 Introduction 1.1 Consider the following equations for the pressure of a real gas. For each equation, find the dimensions of the constants a and b and express these dimensions in SI units. (a) The Dieterici equation: p D RTe.an VRT/.V n/ b Solution: Since an VRT is a power, it is dimensionless and a has the same dimensions as VRT n.
